How does an ip prefix
90.0.0.0/4
translate to
0101∗
in binary
and
10.0.0.0/2
to this:
00*
I'm trying to understand it since a while now but I am unable to derive the logic.
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$\begingroup$ Anything of the form IP/N will translate to "N-bits*", where "N-bits" are the first N bits of the IP (first = most significant). So, in your specific cases you have to take the first 4 bits out of 90, and the first 2 out of 10 (since 90 and 10 convert to an 8 bit byte, and both 4 and 2 are smaller). $\endgroup$– chiMay 16, 2020 at 8:49
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1 Answer
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For a netmasked IP address xxx.xxx.xxx.xxx/N, convert each xxx octet in the address to binary. The masked address is the leftmost N bits with all other bits replaced with zeroes.
So 10.0.0.2/2 becomes 00001010 00000000 00000000 00000010 which is then masked to 00000000 00000000 00000000 00000000, because 00 are the leftmost 2 bits.
90.0.0.0/4 becomes 01011010 00000000 00000000 00000000 which is then masked to 01010000 00000000 00000000 00000000, because 0101 are the leftmost 4 bits.
answered May 16, 2020 at 1:35